What is the weight/weight % of a solution made by dissolving 15.0 g of NaNO_3 in
ID: 998516 • Letter: W
Question
What is the weight/weight % of a solution made by dissolving 15.0 g of NaNO_3 in exactly 100 g of water. 2. What is the molarity of a solution obtained by dissolving 5.35 g of NaBr in some water and diluting to exactly 100 mL in a graduated flask? 3. How many grams of KNO_3 are there in exactly 25.00 mL of a 0.1185 M aqueous solution of the salt? 4. What volume of a 0.3765 M NaCL solution in mL would you measure out to provide exactly 0.2500 g of NaCL? 5. a) Tell what you understand by the term osmosis, and; b) Consider the diagram above with a built in semipermeable membrane as shown: i) Suppose 0.20 M NaCl is placed in arm A and 0.10 M NaCl is placed in arm B both to the levels. What would happen to the liquid levels after equilibrium has been reached? ii) Supposed you had used 0.10 M NaNO_3 in arm A and 0.10 M CaCl_2 in arm B, again both to the same levels initially. What would happen to the levels at equilibrium? iii) Supposed you had used 0.20 M sucrose in arm A and 0.10 M glucrose in arm B again both to the same levels initially. What would happen to the levels at equilibrium?Explanation / Answer
1 ) W/W % = Gram of substance / Gram of solution
= 15 gm / 100 gm
= 0.15 %
2 ) Molarity = No. og moles of solute / Volume in lit
No. of moles of NaBr = g of NaBr / Mol wt of NaBr
= 5.35 / 103
= 0.051 Moles
Therefore Molarity = 0.051 / 0.100 [ 1 Lit = 1000 gm therefore 100 ml = 100 / 1000 = 0.100 lit ]
= 0.516 M
3) Molarity x volume = Gram / Mol wt
Therefore Grams = Molarity x Volume x Mol wt
= 0.1185 x 0.025 x 101 [ 25 mL = 25 / 1000 = 0.025 Lit ]
= 0.299 g of KNO3
4 ) Molarity x volume = Gram / Mol wt
Therefore, Volume = ( Gram x Mol wt ) / Molarity
= ( 0.2500 x 58.5 ) / 0.3765
= 14.625 / 0.3765
= 38.84 mL of NaCl
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